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Cracking the SAT Subject Test in Math 2, 2nd Edition: Everything You Need to Help Score a Perfect 800
Cracking the SAT Subject Test in Math 2, 2nd Edition: Everything You Need to Help Score a Perfect 800
Princeton Review
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EVERYTHING YOU NEED TO HELP SCORE A PERFECT 800. Equip yourself to ace the SAT Subject Test in Math 2 with The Princeton Review's comprehensive study guide—including 3 fulllength practice tests, thorough reviews of key topics, and targeted strategies for every question type.
We don't have to tell you how tough SAT Math is—or how helpful a stellar exam score can be for your chances of getting into your topchoice college. Written by the experts at The Princeton Review, Cracking the SAT Subject Test in Math 2 arms you to take on the test and achieve your highest score.
Techniques That Actually Work.
• Triedandtrue tactics to help you avoid traps and beat the test
• Tips for pacing yourself and guessing logically
• Essential strategies to help you work smarter, not harder
Everything You Need to Know for a High Score.
• Expert subject reviews for every test topic
• Uptodate information on the SAT Subject Test in Math 2
• Score conversion tables to help you assess your performance and track your progress
Practice Your Way to Perfection.
• 3 fulllength practice tests (2 in the book and 1 online) with detailed answer explanations
• Practice drills throughout each content chapter
• Endofchapter summaries to help you master key points
We don't have to tell you how tough SAT Math is—or how helpful a stellar exam score can be for your chances of getting into your topchoice college. Written by the experts at The Princeton Review, Cracking the SAT Subject Test in Math 2 arms you to take on the test and achieve your highest score.
Techniques That Actually Work.
• Triedandtrue tactics to help you avoid traps and beat the test
• Tips for pacing yourself and guessing logically
• Essential strategies to help you work smarter, not harder
Everything You Need to Know for a High Score.
• Expert subject reviews for every test topic
• Uptodate information on the SAT Subject Test in Math 2
• Score conversion tables to help you assess your performance and track your progress
Practice Your Way to Perfection.
• 3 fulllength practice tests (2 in the book and 1 online) with detailed answer explanations
• Practice drills throughout each content chapter
• Endofchapter summaries to help you master key points
سال:
2017
Edition:
2
ناشر کتب:
Princeton Review
زبان:
english
صفحات:
675
ISBN 10:
1524710806
ISBN 13:
9781524710804
سیریز:
College Test Preparation
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Editorial Rob Franek, EditorinChief Casey Cornelius, VP Content Development Mary Beth Garrick, Director of Production Selena Coppock, Managing Editor Meave Shelton, Senior Editor Colleen Day, Editor Sarah Litt, Editor Aaron Riccio, Editor Orion McBean, Associate Editor Penguin Random House Publishing Team Tom Russell, VP, Publisher Alison Stoltzfus, Publishing Director Jake Eldred, Associate Managing Editor Ellen Reed, Production Manager Suzanne Lee, Designer The Princeton Review 555 W. 18th Street New York, NY 10011 Email: editorialsupport@review.com Copyright © 2017 by TPR Education IP Holdings, LLC. Cover art by Stefano Sala / Alamy Stock Photo Cover design by Suzanne Lee All rights reserved. Published in the United States by Penguin Random House LLC, New York, and in Canada by Random House of Canada, a division of Penguin Random House Ltd., Toronto. Terms of Service: The Princeton Review Online Companion Tools (“Student Tools”) for retail books are available for only the two most recent editions of that book. Student Tools may be activated only twice per eligible book purchased for two consecutive 12month periods, for a total of 24 months of access. Activation of Student Tools more than twice per book is in direct violation of these Terms of Service and may result in discontinuation of access to Student Tools Services. Trade Paperback ISBN 9781524710804 Ebook ISBN 9781524710965 SAT Subject Tests is a trademark of the College Board, which is not affiliated with The Princeton Review. The Princeton Review is not affiliated with Princeton University. Editor: Aaron Riccio Production Artist: Deborah A. Silvestrini Production Editor: Harmony Quiroz 2nd Edition v5.2 a Acknowledgments Special thanks to Chris Chimera for his tremendous effort to develop the new content in this edition. Thanks also to Jonathan Chiu, National ACT & SAT Content Director, for his contributions to this title, and to those who have worked on this book in the past. Our gratitude as well to the stellar production team of Deb; orah A. Silvestrini and Harmony Quiroz: they literally make this book look good. Special thanks to Adam Robinson, who conceived of and perfected the Joe Bloggs approach to standardized tests, and many other techniques in the book. Contents Cover Title Page Copyright Acknowledgments Get More (Free) Content Part I: Orientation 1 Introduction 2 Strategy Part II: Diagnostic Practice Test 3 Practice Test 1 4 Practice Test 1: Answers and Explanations Practice Test 1 Answer Key Practice Test 1 Explanations How to Score Practice Test 1 Part III: Content Review 5 Algebra Algebra on the SAT Subject Test in Math 2 Solving Equations Factoring and Distributing Plugging In Plugging In The Answers Inequalities Working with Ranges Direct and Indirect Variation Work and Travel Questions Simultaneous Equations FOIL Method Factoring Quadratics The Quadratic Formula Graphing Calculator to the Rescue! Comprehensive Algebra Drill Summary 6 Fundamentals Definitions Doing Arithmetic Fractions, Decimals, and Percentages Averages Irrational Numbers Exponents Roots Special Exponents More Important Exponent Stuff Comprehensive Fundamentals Drill Summary 7 Plane and Solid Geometry Definitions Plane Geometry Formulas Prisms Rectangular Solids Cubes Cylinders Cones Spheres Pyramids Tricks of the Trade Comprehensive Plane and Solid Geometry Drill Summary 8 Coordinate Geometry Definitions The Coordinate Plane The Equation of a Line Linear Inequalities General Equations Triaxial Coordinates: Thinking in 3D Comprehensive Coordinate Geometry Drill Summary 9 Trigonometry Definitions The Basic Functions Trigonometric Identities Graphing Trigonometric Functions Trigonometry in NonRight Triangles Polar Coordinates Comprehensive Trigonometry Drill Summary 10 Functions Definitions Weird Symbols as Functions Functions Using Standard Notation Compound Functions Inverse Functions Domain and Range Functions Within Intervals: Domain Meets Range Graphing Functions Range and Domain in Graphs Roots of Functions in Graphs Symmetry in Functions Degrees of Functions Comprehensive Functions Drill Summary 11 Statistics and Sets Definitions Working with Statistics Probability Permutations, Combinations, and Factorials Group Questions Comprehensive Statistics and Sets Drill Summary 12 Miscellaneous Logarithms Visual Perception Arithmetic and Geometric Sequences Limits Vectors Logic Imaginary Numbers Polynomial Division Matrices Comprehensive Miscellaneous Drill Summary Part IV: Drills: Answers and Explanations Chapter 5: Algebra Drill Explanations Chapter 6: Fundamentals Drill Explanations Chapter 7: Plane and Solid Geometry Drill Explanations Chapter 8: Coordinate Geometry Drill Explanations Chapter 9: Trigonometry Drill Explanations Chapter 10: Functions Drill Explanations Chapter 11: Statistics and Sets Drill Explanations Chapter 12: Miscellaneous Drill Explanations Part V: Final Practice Test 13 Practice Test 2 14 Practice Test 2: Answers and Explanations Practice Test 2 Answer Key Practice Test 2 Explanations How to Score Practice Test 2 Get More (Free) Content 1 2 3 Go to PrincetonReview.com/cracking. 4 Click the “Student Tools” button, also found under “My Account” from the top toolbar. You’re all set to access your bonus content! Enter the following ISBN for your book: 9781524710965. Answer a few simple questions to set up an exclusive Princeton Review account. (If you already have one, you can just log in.) Need to report a potential content issue? Contact EditorialSupport@review.com. Include: • full title of the book • ISBN number • page number Need to report a technical issue? Contact TPRStudentTech@review.com and provide: • your full name • email address used to register the book • full book title and ISBN • computer OS (Mac/PC) and browser (Firefox, Safari, etc.) Once you’ve registered, you can… • Access a third Math 2 practice test • Take a fulllength practice SAT and/or ACT • Get valuable advice about the college application process, including tips for writing a great essay and where to apply for financial aid • If you’re still choosing between colleges, use our searchable rankings of The Best 382 Colleges to find out more information about your dream school • Access comprehensive study guides and a variety of printable resources, including additional bubble sheets, score conversion tables, and chapter summary pages • Check to see if there have been any corrections or updates to this edition • Get our take on any recent or pending updates to the SAT Subject Test in Math 2 Look For These Icons Throughout The Book ONLINE ARTICLES ONLINE PRACTICE TESTS PROVEN TECHNIQUES APPLIED STRATEGIES COLLEGE ADVISOR APP Part I Orientation 1 2 Introduction Strategy Chapter 1 Introduction Welcome to the SAT Subject Test in Math 2! This chapter will help you get familiar with this book and learn how to use it most effectively. We’ll also talk about when to take the test and how to determine which level to take. (If you’re flipping through this book in the bookstore, this chapter’s for you!) WHAT IS THE SAT SUBJECT TEST IN MATH 2? The SAT Subject Test in Math 2 is a standardized test in mathematics. Colleges use this test to assist in admissions decisions and to place incoming students in classes at the right level. This test is written by ETS, a company in the business of writing tests like these. ETS makes money by charging students to take the SAT and SAT Subject Tests, and charging again to send the scores to colleges. You’ll also run into ETS exams if you ever apply to graduate school. The SAT Subject Test in Math 2 has 50 multiplechoice questions and is one hour long. The test is scored from 200 to 800 points. The SAT Subject Test in Math 2 covers a range of mathematical topics, from basic algebra to trigonometry and statistics. Many colleges require some SAT Subject Tests (frequently two, but occasionally one or three). The subjects available are varied: two in mathematics, three in science, two in history, one in English, and twelve in foreign languages. Different schools have different preferences and requirements for which tests to take, too. For example, an engineering program may want to see one math and one science. Check each school’s website to determine how many tests you must take and which ones (if any) are preferred. What’s on the Test? The content of the SAT Subject Test in Math 2 is approximately as follows: Topic Math Level 2 Functions 12 questions Trigonometry 10 questions Algebra 9 questions Coordinate Geometry 6 questions Solid Geometry 3 questions Statistics 4 questions Miscellaneous 6 questions TOTAL 50 questions As you can see, the SAT Subject Test in Math 2 focuses on material you learned in your Geometry, Algebra II, and Precalculus classes. When it asks questions about basic concepts, it does so by including the concepts in a more complicated problem. For example, there are no direct questions about plane geometry. However, you will need to be able to apply the concepts of plane geometry to questions about coordinate geometry or spatial geometry. You may be overwhelmed by the number of different topics which appear on the SAT Subject Test in Math 2. Fear not! The test is written with the expectation that most students have not covered all the material on the test. Furthermore, you can do well on this test even if you haven’t covered everything that may show up on the test. Math Level 1 or 2? We’d love to say that this decision boils down completely to your comfort level with the material in each course, but the truth is that not every school accepts the Math 1 results. You should therefore base your decision primarily on the admission requirements of the schools that interest you. Math 2 is appropriate for high school students who have had a year of trigonometry or precalculus and have done well in the class. You should also be comfortable using a scientific or graphing calculator. If you hate math, do poorly on math tests, or have not yet studied Trigonometry or Precalculus, the SAT Subject Test in Math 2 is probably not for you. It’s worth noting, however, that while this test is difficult, the test is scored on a comparatively generous curve. If you find yourself making random (or “silly”) mistakes more than anything else, the Math 2 scoring grid may work in your favor. Colleges also receive your percentile (comparing you to other test takers), as well as your scaled (200–800) score. For the most part, they pay attention to the scaled score and ignore the percentile. However, to the small extent that percentiles matter, Math 1 has considerably more forgiving percentiles. People who take Math 2 are generally really good at math; about 13% of them get a perfect score! Less than 1% of Math 1 testtakers get a perfect score, though. As a result, a 790 on Math 2 is only in the 85th percentile (about 13% get an 800 and 2% get a 790), while a 790 on Math 1 is still in the 99th percentile. The disparity between the percentiles continues down the entire score range. If you are very unsure about which test to take, even after working practice questions and taking practice tests, you can take both tests. Want to know which colleges are best for you? Check out The Princeton Review’s College Advisor app to build your ideal college list and find your perfect college fit! Available for free in the iOS App Store and Google Play Store. WHEN SHOULD I TAKE THE SAT SUBJECT TEST IN MATH 2? The right time to take the SAT Subject Test in Math 2 varies from person to person. Many students take the test at the end of a Precalculus class in school. (Precalculus also goes by many other names, such as Trigonometry, Advanced Functions, or other less recognizable names.) Some students take Math 2 during or at the end of an AP Calculus course. The SAT Subject Tests are offered six times per year, and no test date is easier or harder than any other test date. The most popular test dates are in May and June, because these are at the end of a school year when the material is freshest in the student’s mind. Whenever you choose to take the test, make sure you have time to do some practice beforehand, so that you can do your best (and not have to take the thing again!). The Calculator The SAT Subject Test in Math 2 is designed to be taken with the aid of a calculator. Students taking this test should have a scientific or graphing calculator and know how to use it. A “scientific” calculator is one that has keys for the following functions: • the values of π and e • square roots • raising to an exponent • sine, cosine, and tangent • logarithms Calculators without these functions will not be as useful. Graphing calculators are allowed. The graphing features on a graphing calculator are helpful on a fairly small number of questions per test, and they are necessary for about 0–1 questions per test. If you’re going to take a graphing calculator to the test, make sure you know how to use it. Fumbling over your calculator trying to figure something out during the test is just not a productive use of your time! This book is going to focus on the TI84. If you have another family member of the TI80 series, know that these comments still apply to you with minor adjustments. Check with your manual for specific key stroke changes. If you have a scientific calculator, we’ll be showing you your key stroke changes in the sidebars throughout the manual. The ETS Predictor ETS says that for the SAT Subject Test in Math 2, a calculator may be useful or necessary for about 5565 percent of the questions. Certain kinds of calculators are not allowed. For example, a calculator with a QWERTY keyboard (like a computer keyboard) is not allowed. Your calculator must not require a wall outlet for power and must not make noise or produce paper printouts. There will be no replacements at the test center for malfunctioning or forgotten calculators, though you’re welcome to take a spare, as well as spare batteries. Laptops, tablets, and cell phones are also not allowed as calculators. Bottom line: You need a calculator for this test. Certain things will be easier with a graphing calculator, but it is most important that you are comfortable using your calculator. HOW TO USE THIS BOOK It’s best to work through the chapters of this book in sequence, since the later chapters build on the techniques introduced in earlier chapters. If you want an overall review of the material on the SAT Subject Test in Math 2, just start at the beginning and cruise through to the end. This book will give you all the techniques and knowledge you need to do well on the test. If you feel a little shaky in certain areas of math and want to review specific topics, the chapter headings and subheadings will also allow you to zero in on your own problem topics. As with any subject, pay particular attention to the math topics you don’t like—otherwise, those are the ones that will burn you on the real test. Study Guide In your free online student tools, you’ll find a printable copy of two different study guides that you can use to plot out your progress through this book. If you really want to get your money’s worth out of this book, you’ll follow this study plan. • Take the diagnostic fulllength practice test that appears in Part II. • Score your test and review it to see where your strengths and weaknesses lie. • Focus on the chapters associated with the questions that you got wrong or didn’t understand. • Read through each section of a chapter carefully until you feel that you understand it. • Try the practice questions at the end of that section. • Check your answers, and review any questions you got wrong until you understand your mistakes. • Once you finish all the sections in a chapter, try the Comprehensive Drill at the end of that chapter. • Check your answers, and review any questions you got wrong until you understand your mistakes. • Go back and review the chapters which cover material you’re still struggling with. • Take the final inbook fulllength practice test. Then score and review it. If you put in the work and study what’s in this book, you’ll be prepared for anything that ETS may throw at you on the day of the real test. Need More? You can also visit CollegeBoard.org for more information and test questions. Also, don’t forget to register your book online for access to a third fulllength practice test! Chapter 2 Strategy It’s easy to get the impression that the only way to excel on the SAT Subject Test in Math 2 is to become an expert on myriad mathematical matters. However, there are many effective strategies that you can use. From Pacing to Process of Elimination to using your calculator, this chapter takes you through the most important general strategies so you can start practicing them right way. CRACKING THE SAT SUBJECT TEST IN MATH 2 It’s true that you have to know some math to do well, but there’s a great deal you can do to improve your score without staring into math books until you go blind. Several important strategies will help you increase your scoring power. • The questions on the SAT Subject Test in Math 2 are arranged in order of difficulty. You can think of a test as being divided roughly into thirds, containing easy, medium, and difficult questions, in that order. • The SAT Subject Test in Math 2 is a multiplechoice test. That means that every time you look at a question on the test, the correct answer is on the paper right in front of you. • ETS writes incorrect answers on the SAT Subject Test in Math 2 by studying errors commonly made by students. These are common errors that you can learn to recognize. The next few pages will introduce you to testtaking techniques that use these features of the SAT Subject Test in Math 2 to your advantage, which will increase your score. These strategies come in two basic types: Section strategies (which help you determine which questions to do and how much time to spend on them) and question strategies (which help you solve an individual question once you’ve chosen to do it.) SECTION STRATEGY The following represents a sample scoring grid from the SAT Subject Test in Math 2. Note that scoring scales will vary from test to test, so this is just a general guide. Raw Score Scaled Score Percentile 43–50 800 87 42 790 85 41 780 82 40 770 79 39 760 77 38 750 73 37 740 71 36 730 68 35 720 66 34 710 62 33 700 61 31–32 690 58 30 680 56 29 670 53 28 660 50 27 650 47 26 640 44 24–25 630 40 23 620 37 22 610 33 21 600 31 19–20 590 28 18 580 25 17 570 22 15–16 560 20 — 550 18 14 540 15 13 530 14 12 520 12 — 510 10 11 500 8 10 490 7 — 480 6 9 470 4 8 460 3 — 450 3 7 440 2 6 430 2 — 420 1 — 410 1 4–5 400 1 — 390 1 3 380 1 2 370 1 1 360 1 A few things are notable: • While it’s theoretically possible to score less than a 350, to do so would require you to score a negative number of raw points (i.e., do worse than simply randomly guessing). Practically speaking, the scoring range on the SAT Subject Test in Math 2 is from 350–800. • On some test dates, some scores are not possible, such as 420 in the test shown above. • The scoring grid for the SAT Subject Test in Math 2 is very forgiving, especially at the top end. Anything from 43 to 50 raw points gets you a “perfect” 800, and 33 raw points out of a possible 50 is still a 700. However, the percentiles are brutal: a 700 is only the 61st percentile! Pacing The first step to improving your performance on the SAT Subject Test in Math 2 is slowing down. That’s right: You’ll score better if you do fewer questions. It may sound strange, but it works. That’s because the testtaking habits you’ve developed in high school are poorly suited to the SAT Subject Test in Math 2. It’s a different kind of test. One Point Over Another? A hard question on the SAT Subject Test in Math 2 isn’t worth more points than an easy question. It just takes longer to do, and it’s harder to get right. It makes no sense to rush through a test if all that’s waiting for you are tougher and tougher questions—especially if rushing worsens your performance on the easy questions. Think about a freeresponse math test. If you work a question and get the wrong answer, but you do most of the question right, show your work, and make a mistake that lots of other students in the class make (so the grader can easily recognize it), you’ll probably get partial credit. If you do the same thing on the SAT Subject Test in Math 2, you get one of the four wrong answers. But you don’t get partial credit for choosing one of the listed wrong answers; you lose a quarterpoint. That’s the opposite of partial credit! Because the SAT Subject Test in Math 2 gives the opposite of partial credit, there is a huge premium on accuracy in this test. How Many Questions Should I Do? Use the following chart to determine how many questions you should attempt the next time you take an SAT Subject Test in Math 2: As you improve, your pacing goals will also get more aggressive. Once you take your next practice test and score it, come back to this chart and adjust your pacing accordingly. For example, if you initially scored a 550, but on your second test you scored a 610, then use the 610–650 line for your third test, and you may score a 700 (or even higher!). Your Last Test For your “last test,” use your last SAT Subject Test in Math 2 (real or practice), if you’ve taken one. If you’ve taken the SAT, use your Math score. You can also use a PSAT score; just add a “0” to the end of your Math score (so a 56 becomes a 560). If you’ve taken the ACT instead, multiply your math score by 20 (so a 25 in Math becomes a 500 for the purpose of pacing on the SAT Subject Test in Math 2). If you haven’t taken any of these tests, make an educated guess! Personal Order of Difficulty You probably noticed that the previous chart doesn’t tell you which questions to do on the SAT Subject Test in Math 2, only how many. That’s because students aren’t all the same. Even if a certain question is easy for most students, if you don’t know how to do it, it’s hard for you. Conversely, if a question is hard for most students but you see exactly how to do it, it’s easy for you. Most of the time, you’ll find lowernumbered questions easy for you and highernumbered questions harder for you, but not always, and you should always listen to your Personal Order of Difficulty (POOD). Develop a Pacing Plan The following is an example of an aggressive pacing plan. You should begin by trying this plan, and then you should adapt it to your own needs. First, do questions 1–20 in 20 minutes. They are mostly easy, and you should be able to do each one in about a minute. (As noted above, though, you must not go so quickly that you sacrifice accuracy.) If there is a question that looks more timeconsuming, but you know how to do it, mark it so that you can come back to it later, but move on. Second, pick and choose among questions 21–50. Do only questions that you are sure you can get right quickly. Mark questions that are more timeconsuming (but you still know how to do them!) so that you can come back to them later. Cross out questions that you do not know how to do; you shouldn’t waste any more time on them. Third, once you’ve seen every question on the test at least once and gotten all the quick points that you can get, go back to the more timeconsuming questions. Make good choices about which questions to do; at this point, you will be low on time and need to make realistic decisions about which questions you will be able to finish and which questions you should give up for lost. This pacing plan takes advantage of the test’s builtin order of difficulty and your POOD. You should move at a brisk but not breakneck pace through the easy questions so that you have enough time to get them right but not waste time. You should make sure that you get to the end of the test and evaluate every question, because you never know if you happen to know how to do question 50; it may be harder for most students than question 30, but it just may test a math topic that you remember very well from class (or this book). Delaying more timeconsuming questions until after you’ve gotten the quick and easy points maximizes your score and gives you a better sense of how long you have to complete those longer questions, and, after some practice, it will take only a few seconds to recognize a timeconsuming question. A Note About Question Numbers As you cruise through this book, you’ll run into practice questions that seem to be numbered out of order. That’s because the numbers of the practice questions tell you what position those questions would occupy on a 50question SAT Subject Test in Math 2. The question number gives you an idea of how difficult ETS considers a given question to be. QUESTION STRATEGY It’s true that the math on the SAT Subject Test in Math 2 gets difficult. But what exactly does that mean? Well, it doesn’t mean that you’ll be doing 20step calculations, or huge, crazy exponential expansions that your calculator can’t handle. Difficult questions on this test require you to understand some slippery mathematical concepts, and sometimes to recognize familiar math rules in strange situations. This means that if you find yourself doing a 20step calculation, stop. There’s a shortcut, and it probably involves using one of our techniques. Find it. Random Guessing If you randomly guess on five questions, you can expect to get one right and four wrong. Your score for those five questions will be: . This isn’t very helpful. Process of Elimination (POE) It’s helpful that the SAT Subject Test in Math 2 contains only multiplechoice questions. After all, this means that eliminating four answers that cannot possibly be right is just as good as knowing what the right answer is, and it’s often easier. Eliminating four answers and choosing the fifth is called the Process of Elimination (POE). POE Guessing If you correctly eliminate two answer choices and guess among the remaining three, you have a oneinthree chance of getting the right answer. If you do this on six questions, you can expect to get two right and four wrong. Your score will be : . That’s not a lot for six questions, but every point helps. POE can also be helpful even when you can’t get down to a single answer. Because of the way the test is scored (plus one raw point for a correct answer and minus a quarterpoint for an incorrect answer), if you can eliminate at least one answer, it is to your advantage to guess. So, the bottom line: To increase your score on the SAT Subject Test in Math 2, eliminate wrong answer choices whenever possible, and guess aggressively whenever you can eliminate anything. There is a major elimination technique you should rely on as you move through the test: ballparking. Ballparking Sometimes, you can approximate an answer: You can eliminate answer choices by ballparking whenever you have a general idea of the correct answer. Answer choices that aren’t even in the right ballpark can be crossed out. Take a look at the following three questions. In each question, at least one answer choice can be eliminated by ballparking. See whether you can make eliminations yourself. For now, don’t worry about how to do these questions—just concentrate on eliminating answer choices. 6. If = 1.84, then x2 = (A) −10.40 (B) −3.74 (C) 7.63 (D) 10.40 (E) 21.15 Here’s How to Crack It You may not have been sure how to work with that ugly fractional exponent. But if you realized that x2 can’t be negative, no matter what x is, then you could eliminate (A) and (B)—the negative answers—and then guess from the remaining answer choices. Figure 1 13. In Figure 1, if c = 7 and θ = 42°, what is the value of a ? (A) 0.3 (B) 1.2 (C) 4.7 (D) 5.2 (E) 6.0 Here’s How to Crack It Unless you’re told otherwise, the figures that the SAT Subject Test in Math 2 gives you are drawn accurately, and you can use them to ballpark. In this example, even if you weren’t sure how to apply trigonometric functions to the triangle, you could still approximate based on the diagram provided. If c is 7, then a looks like, say, 5. That’s not specific enough to let you decide between (C), (D), and (E), but you can eliminate (A) and (B). They’re not even close to 5. At the very least, that gets you down to a 1in3 guess—much better odds. Can I Trust The Figure? In order to intentionally mislead you, sometimes ETS inserts figures that are deliberately inaccurate. When the figure is wrong, ETS will print underneath, “Note: Figure not drawn to scale.” When you see this note, trust the text of the problem, but don’t believe the figure, because the figure is just there to trick you. 22. The average (arithmetic mean) cost of Simon’s math textbooks was $55.00, and the average cost of his history textbooks was $65.00. If Simon bought 3 math textbooks and 2 history textbooks, what was the average cost of the 5 textbooks? (A) $57.00 (B) $59.00 (C) $60.00 (D) $63.50 (E) $67.00 Here’s How to Crack It Here, once again, you might not be sure how to relate all those averages. However, you could realize that the average value of a group can’t be bigger than the value of the biggest member of the group, so you could eliminate (E). You might also realize that, since there are more $55 books than $65 books, the average must be closer to $55.00 than to $65.00, so you could eliminate (C) and (D). That gets you down to only two answer choices, a 50/50 chance. Those are excellent odds. These are all fairly basic questions. By the time you’ve finished this book, you won’t need to rely on ballparking to answer them. The technique of ballparking will still work for you, however, whenever you’re looking for an answer you can’t figure out with actual math. “Better” Than Average What makes a question hard? Sometimes, a hard question tests more advanced material. For example, on the SAT Subject Test in Math 2, questions about polar coordinates are rare before question 20. Sometimes a hard question requires more steps, four or five rather than one or two. But more often, a hard question has trickier wording and better trap answers than an easy question. ETS designs its test around certain trends and traps, looking to catch the average student with the sort of tricks and problems that have tripped testtakers up in the past. While this does mean that you’ll have to be alert, it also means that many of these questions have predictable wrong answers, and you can use this knowledge to “beat” the curve. When ETS writes a question that mentions “a number,” it counts on students to think of numbers like 2 or 3, not numbers like −44.76 or 4π. ETS counts on students to think of the most obvious thing, like 2 or 3 instead of −44.76 or 4π. Don’t be led astray by the urge to choose these; instead, use it to your advantage. There is nothing on the SAT Subject Test in Math 2 that hasn’t been taught to students, which means that in order to trip students up, the test writers need to make students pick a wrong answer. It does that by offering answers that are too good to be true: tempting oversimplifications, obvious answers to subtle questions, and all sorts of other answers that seem comforting and familiar. Take a step back. Try eliminating choices like these and then pick and check another one instead. 28. Ramona cycles from her house to school at 15 miles per hour. Upon arriving, she realizes that it is Saturday and immediately cycles home at 25 miles per hour. If the entire roundtrip takes her 32 minutes, then what is her average speed, in miles per hour, for the entire roundtrip? (A) 17.0 (B) 18.75 (C) 20.0 (D) 21.25 (E) 22.0 Here’s How to Crack It This is a tricky problem, and you may not be sure how to solve it. You can, however, see that there’s a very tempting answer among the answer choices. If someone goes somewhere at 15 mph and returns at 25 mph, then it seems reasonable that the average speed for the trip should be 20 mph. For question 28, however, that’s far too obvious to be right. Eliminate (C). Stop and Think Anytime you find an answer choice immediately appealing on a hard question, stop and think again. ETS collects data from thousands of students in trial tests before making a question a scored part of their tests. If it looks that good to you, it probably looked good to many of the students taking the trial tests. That attractive answer choice is almost certainly a trap. The right answer won’t be the answer most people would pick. On hard questions, obvious answers are wrong. Eliminate them. 34. If θ represents an angle such that sin2θ = tanθ − cos2θ, then sinθ − cosθ = (A) − (B) 0 (C) 1 (D) 2 (E) It cannot be determined from the information given. Here’s How to Crack It On a question like this one, you might have no idea how to go about finding the answer. That “It cannot be determined” answer choice may look awfully tempting. You can be sure, however, that (E) will look tempting to many students. It’s too tempting to be right on a question this hard. You can eliminate (E). 48. If the above cones are similar, and the volume of the larger cone is 64, then what is the volume of the smaller cone? (A) 2 (B) 4 (C) 8 (D) 16 (E) 32 Here’s How to Crack It This one may seem simple: the smaller cone is half as tall as the larger cone, so its volume must be = 32. But wait! This is question number 48. That means that most test takers will miss it. We’ll cover how to tackle this question easily in the Solid Geometry chapter, but before you turn the page, be sure to cross out 32, as it's a trap answer! SO DO I HAVE TO KNOW MATH AT ALL? The techniques in this book will go a long way toward increasing your score, but there’s a certain minimum amount of mathematical knowledge you’ll need in order to do well on the SAT Subject Test in Math 2. We’ve collected the most important rules and formulas into lists. As you move through the book, you’ll find these lists at the end of each chapter. The strategies in this chapter, and the techniques in the rest of this book, are powerful tools. They will make you a better test taker and improve your performance. Nevertheless, memorizing the formulas on our lists is as important as learning techniques. Memorize those rules and formulas, and make sure you understand them. Using That Calculator Behold the First Rule of Intelligent Calculator Use: Your calculator is only as smart as you are. It’s worth remembering. Some test takers have a dangerous tendency to rely too much on their calculators. They try to use them on every question and start punching numbers in even before they’ve finished reading a question. That’s a good way to make a question take twice as long as it has to. The most important part of problem solving is done in your head. You need to read a question, decide which techniques will be helpful in answering it, and set up the question. Using a calculator before you really need to do so will keep you from seeing the shortcut solution to a problem. Scientific or Graphing? ETS says that the SAT Subject Test in Math 2 is designed with the assumption that most test takers have graphing calculators. ETS also says that a graphing calculator may give you an advantage on a handful of questions. If you have access to a graphing calculator and know how to use it, you may want to choose it instead of a scientific calculator. When you do use your calculator, follow these simple procedures to avoid the most common calculator errors. • Check your calculator’s operating manual to make sure that you know how to use all of your calculator’s scientific functions (such as the exponent and trigonometric functions). • Clear the calculator at the beginning of each problem to make sure it’s not still holding information from a previous calculation. • Whenever possible, do long calculations one step at a time. It makes errors easier to catch. • Write out your work! Label everything, and write down the steps in your solution after each calculation. That way, if you get stuck, you won’t need to do the entire problem over again. Writing things down will also prevent you from making careless errors. • Keep an eye on the answer choices to see if ETS has included a partial answer designed to tempt you away from the final answer. Eliminate it! Above all, remember that your brain is your main problemsolving tool. Your calculator is useful only when you’ve figured out exactly what you need to do to solve a problem. Set It Up! Some questions on the SAT Subject Test in Math 2 can be answered without much calculation—the setup itself makes the answer clear. Remember: Figure out how to do the problem with your brain; then do the problem with your calculator. Part II Diagnostic Practice Test 3 4 Practice Test 1 Practice Test 1: Answers and Explanations Chapter 3 Practice Test 1 Click here to download and print the PDF version of this exercise. MATHEMATICS LEVEL 2 For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding oval on the answer sheet. Notes: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions on this test. For each question, you will have to decide whether or not you should use a calculator. (2) The only angle measure used on this test is degree measure. Make sure that your calculator is in degree mode. (3) Figures that accompany problems on this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f(x) is a real number. The range of f is assumed to be the set of all real numbers f(x), where x is in the domain of f. (5) Reference information that may be useful in answering the questions on this test can be found below. THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME OF THE QUESTIONS ON THIS TEST. Volume of a right circular cone with radius r and height h: V = πr2h Lateral area of a right circular cone with circumference of the base c and slant height ℓ: S = cℓ Volume of a sphere with radius r: V = πr3 Surface area of a sphere with radius r: S = 4πr2 Volume of a pyramid with base area B and height h: V = Bh 1. If 2y + 6 = (y + 3) for all y, then c = (A) (B) 2 (C) 9 (D) 15 (E) 18 2. The relationship between a temperature F in degrees Fahrenheit and a temperature C in degrees Celsius is defined by the equation F = C + 32, and the relationship between a temperature in degrees Fahrenheit and a temperature R in degrees Rankine is defined by the equation R = F + 460. Which of the following expresses the relationship between temperatures in degree Rankine and degrees Celsius? (A) R = C − 32 + 460 (B) R = C + 32 + 460 (C) R = C + 32 − 460 (D) R = C + 860 (E) R = C − 828 3. What is the slope of a line containing the points (1, 13) and (−3, 6)? (A) 0.14 (B) 0.57 (C) 1.75 (D) 1.83 (E) 6 4. If a + b + c = 12, a +b = 4, and a + c = 7, what is the value of a ? (A) 2 (B) 1 (C) (D) 2 (E) 5. If g(x) = 2ex − 2 and h(x) = ln(x), then g(h(7)) = (A) 7.69 (B) 12 (C) 14 (D) 26.43 (E) 31.98 6. The intersection of a cylinder and a plane could be which of the following? I. A circle II. A triangle III. A rectangle (A) I only (B) II only (C) I and III only (D) II and III only (E) I, II, and III 7. The figure above shows a helium balloon rising vertically. When the balloon reaches a height of 54 inches, the angles of elevation from points X and Y on the ground are 72.4° and 50.8°, respectively. What is the distance, in inches, between points X and Y ? (A) 61.17 (B) 72.29 (C) 84.15 (D) 124.72 (E) 236.44 8. What is the value of y2 if ? (A) 2562 (B) 256 (C) 16 (D) 4 (E) 2 9. The points in the xyplane are transformed so that each point A(x, y) is transformed to A’ (3x, 3y). If the distance between point A and the origin is c, then the distance between the point A’ and the origin is (A) (B) (C) c (D) c (E) 3c 10. If (A) x2 − 2 and , then q(x) = (B) x2 (C) x (D) (E) 11. If x is the degree measure of an angle such that 0° < x < 90° and cosx = 0.6, then sin (90° − x) = (A) 0.4 (B) 0.5 (C) 0.6 (D) 0.7 (E) 0.8 12. The set of points defined by the equation x2 + y2 + z2 = 4 is (A) a point (B) a line (C) a circle (D) a plane (E) a sphere 13. The graph of the function g, where vertical asymptote at x = (A) 0 only (B) 3 only (C) 7 only (D) 0 and 3 only (E) 0, 3, and 7 , has a 14. The graph of y = x4 + 8x3 − 4x2 − 64x + k is shown above. Which of the following could be the value of k? (A) 1,240 (B) 520 (C) 14 (D) −14 (E) −1,240 15. If sinx = 0.6743, then cscx = (A) 0.6481 (B) 0.8374 (C) 1.2953 (D) 1.4830 (E) 1.9637 16. Sarah is planning a vacation at a hotel that costs $80 per night. Sarah must also pay the $170 airfare to get there and will also pay for an equally priced hotel room for a friend who will be visiting her on three of the nights. Which of the following correctly expresses the average cost, in dollars, for each night as a function of n, the number of nights of the vacation? (A) (B) (C) (D) (E) 17. Which of the following is an equation whose graph is a set of points equidistant from the points (0, 0) and (6, 0)? (A) x = 3 (B) y = 3 (C) x = 3y (D) y = 3x (E) y = 3x + 3 18. What is the sum of the infinite geometric series (A) (B) (C) (D) 1 (E) 19. Which of the following is equivalent to a− b ≥ a + b? (A) a ≤ b (B) a ≤ 0 (C) b ≤ a (D) b ≤ 0 (E) b ≥ 0 20. If m and n are in the domain of a function g and g(m) > g(n), which of the following must be true? (A) mn ≠ 0 (B) m > n (C) m < n (D) m = n (E) m ≠ n 21. In a certain office, the human resources department reports that 60% of the employees in the office commute over an hour on average each day, and that 25% of those employees who commute over an hour on average each day commute by train. If an employee at the office is selected at random, what is the probability that the employee commutes over an hour on average by train? (A) 0.10 (B) 0.15 (C) 0.20 (D) 0.25 (E) 0.30 22. To the nearest degree, what is the measure of the second smallest angle in a right triangle with sides 5, 12, and 13 ? (A) 23 (B) 45 (C) 47 (D) 60 (E) 67 23. Which of the following is an equation of a line perpendicular to y = 3x − 5 ? (A) y = 5x − 3 (B) y = −3x + 5 (C) y = x + 5 (D) y = − x + 4 (E) 24. What is the range of the function g(x) = −2 + 5cos (3x + 7π) ? (A) −1 ≤ g(x) ≤ 1 (B) −5 ≤ g(x) ≤ −1 (C) −5 ≤ g(x) ≤ 5 (D) −7 ≤ g(x) ≤ 3 (E) −7 ≤ g(x) ≤ 5 25. Of the following list of numbers, which has the greatest standard deviation? (A) 1, 2, 3 (B) 2, 2, 2 (C) 2, 4, 6 (D) 4, 7, 10 (E) 6, 8, 10 26. The formula F =Ie0.06y gives the final amount F that a bank account will contain if an initial investment I is compounded continuously at an annual interest of 6% for y years. Using this formula, after how many years will an initial investment of $100 be worth approximately $600? (A) 5.2 (B) 6.0 (C) 13.0 (D) 22.4 (E) 29.7 27. If cosθ < 0 and > 0, then θ must be in which quadrant in the figure above? (A) I (B) II (C) III (D) IV (E) There is no quadrant in which both conditions are true. 28. If g(−x) = −g(x) for all real numbers x and if (4, 9) is a point on the graph of g, which of the following points must also be on the graph of g ? (A) (−9, −4) (B) (−4, −9) (C) (−4, 9) (D) (4, −9) (E) (9, 4) If a is a multiple of 10, then a is a multiple of 5. 29. If a is an integer, which of the following CANNOT be inferred from the statement above? (A) If a is a multiple of 5, then a is a multiple of 10. (B) If a is not a multiple of 5, then a is not a multiple of 10. (C) a is a multiple of 10 implies that a is a multiple of 5. (D) A necessary condition for a to be a multiple of 10 is that a is a multiple of 5. (E) In order for a to be a multiple of 5, it is sufficient that a be a multiple of 10. 30. In how many different orders can 8 different colors of flowers be arranged in a straight line? (A) 8 (B) 64 (C) 40,320 (D) 80,640 (E) 16,777,216 31. What value does (A) 0 (B) 0.5 (C) 1 approach as x approaches 0 ? (D) 2 (E) It does not approach a unique value 32. If f(x) = 7 − 5x, then f(1) = (A) f(1) (B) f(0) (C) (D) f(2) (E) 33. What is the period of the graph of y = 3tan (2πx + 9) ? (A) (B) (C) 3 (D) (E) 34. The figure above shows a map of Maple Street and Elm Street. Katherine is biking from Point X to Point Y. The straightline distance from Point X to Point Y is 40 kilometers. If Katherine bikes at an average speed of 15 km per hour along Maple Street and Elm Street, how long will it take Katherine to get to Point Y ? (A) 40 minutes (B) 2 hours and 35 minutes (C) 2 hours and 40 minutes (D) 3 hours and 15 minutes (E) 3 hours and 35 minutes x g(x) −2 0 −1 −3 0 2 1 0 2 0 35. If g is a polynomial of degree 4, five of whose values are shown in the table above, then g(x) could equal (x + 1)(x + 2)2 (A) g(x) = (B) g(x) = (x − 2)(x − 1)(x + 2)(x + 3) (C) g(x) = (x − 2) (x + 1)(x + 2) (D) g(x) = (x − 3)(x − 2)(x − 1)(x + 2) (E) g(x) = (x − 2)(x − 1) (x + 2) 36. The only prime factors of an integer m are 2, 3, 5, and 13. Which of the following could NOT be a factor of m ? (A) 6 (B) 9 (C) 12 (D) 26 (E) 35 37. If 0 ≤ x ≤ and cosx = 4sinx, what is the value of x ? (A) 0.245 (B) 0.250 (C) 0.328 (D) 1.217 (E) 1.326 38. If (A) 0.04 (B) 1.73 (C) 3.17 (D) 5.00 (E) 25.98 , what is the value of g−1(15) ? 39. The Triangular Number Sequence Tn can be defined recursively as T1 = 1 Tn = Tn − 1 + n for n > 1 What is the 11th term of the sequence? (A) 45 (B) 55 (C) 66 (D) 78 (E) 91 40. If f(x) = x3 + x2 − 16x + 12, which of the following statements are true? I. The equation f(x) = 0 has three real solutions II. f(x) ≥ −8 for all x ≥ 0 III. The function is increasing for x > 2 (A) I only (B) III only (C) I and III only (D) II and III only (E) I, II, and III only 41. Portions of the graphs of g and h are shown above. Which of the following could be a portion of the graph of gh ? (A) (B) (C) (D) (E) 42. The set of all real numbers y such that is (A) all real numbers (B) no real numbers (C) negative real numbers only (D) nonnegative real numbers only (E) zero only 43. In the triangle shown above, sinx = (A) (B) (C) (D) (E) 44. The length, width, and height of a rectangular solid are 6, 3, and 2. What is the length of the longest segment that can be drawn between two vertices of the solid? (A) 6 (B) 3 (C) 7 (D) 12 (E) 18 45. If logn2 = a and logn5 = b, then logn50 = (A) a + b (B) a + b2 (C) ab2 (D) a + 2b (E) a + 5b 46. If cosx = a, then, for all x, in the interval 0 < x < , tanx = (A) a2 + 1 (B) (C) (D) (E) 47. Which of the following shifts in the graph of y = x2 would result in the graph of y = x2 + 4x + c, where c is a constant greater than 5? (A) Left 2 units and up c − 4 units (B) Right 2 units and down c − 4 units (C) Right 2 units and down c + 4 units (D) Left 2 units and up c + 4 units (E) Right 4 units and up c units 48. If the height of a right square pyramid is increased by 12%, by what percent must the side of the base be increased, so that the volume of the pyramid is increased by 28%? (A) 3% (B) 7% (C) 10% (D) 36% (E) 56% 49. If Matrix X has dimensions a ×b and Matrix Y has dimensions b × c, where a, b, and c are distinct positive integers, which of the following must be true? I. The product XY exists and has dimensions a × c. II. The product XY exists and has dimensions b × b. III. The product YX does not exist. (A) I only (B) II only (C) III only (D) I and III only (E) II and III only 50. If z is the complex number shown in the figure above, which of the following could be iz? (A) A (B) B (C) C (D) D (E) E STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT WORK ON ANY OTHER TEST IN THIS BOOK. Chapter 4 Practice Test 1: Answers and Explanations PRACTICE TEST 1 ANSWER KEY PRACTICE TEST 1 EXPLANATIONS 1. E The question asks for the value of c in an equation that is true for all values of y, so plug in a value for y. To make the math easy on the right side, plug in y = 6. Substitute y = 6 into the equation to get 2(6) + 6 = (6 + 3). Simplify to get 18 = (9). Cancel the 9s on the right side to get 18 = c. The correct answer is (E). 2. B The question asks for the relationship between R and C. Since there are variables in the choices, Plug In. Let C = 10. If C = 10, then F = (10) + 32 = 50. If F = 50, then R = 50 + 460 = 510. Plug C = 10 and R = 510 into each choice and eliminate any that aren’t true. Choice (A) is 510 = (10) − 32 + 460. This is false, so eliminate (A). Choice (B) is 510 = (10) + 32 + 460. This is true, so keep (B). Choice (C) is 510 = (10) + 32 − 460. This is false, so eliminate (C). Choice (D) is 510 = (10) + 860. This is false, so eliminate (D). Choice (E) is 510 = (10) − 828. This is false, so eliminate (E). The correct answer is (B). 3. C To find the slope of a line, use the slope formula: slope = . Let (x1, y1) = (1, 13) and (x2, y2) = (−3, 6). The slope of the line is 4. B . The correct answer is (C). The question includes three equations and three variables, so find a way to combine the equations. Because the second equation provides a value for a + b, substitute this value into the equation for a + b + c to get the value of c. When a + b = 4 is substituted into a + b +c = 12, the result is 4 + c = 12. Subtract 4 from both sides to get c = 8. Now, substitute c = 8 into a + c = 7 to get a + 8 = 7. Subtract 8 from both sides to get a = −1. The correct answer is (B). 5. B The question asks for g(h(7)). On questions involving composition of functions, start on the inside and work toward the outside. Find h(7). Since h(x) = ln(x), h(7) = ln(7) ≈ 1.94591015. To find g(h(7)), find g(1.94591015). Since g(x) = 2ex − 2, g(1.94591015) = 2e1.94591015 − 2 = 12. The correct answer is (B). 6. C The question asks for which of the three listed figures could be the intersection of a plane and a cylinder. Go through one at a time. For (I), since the bases of a cylinder are circles, the plane could intersect the cylinder in a way that the plane contains one of the bases and forms a circle. Since the intersection could form a circle, (I) must be included. Eliminate (B) and (D), which don’t include (I). Try (II). There doesn’t seem to be an obvious way to form a triangle. However, don’t eliminate (II) right away in case there is a way that isn’t obvious. Try (III). Determine whether a rectangle can be formed. If the plane passes through the diameters of each base, then a rectangle is formed. Therefore, (III) must be included, so eliminate (A). Now, come back to (II). If the plane is parallel to the bases, a circle is formed rather than a triangle. If the plane is perpendicular to the bases, a rectangle is formed. If the plane is at any other angle, a curved path is formed, so the result cannot be a triangle. Therefore, eliminate (E), which includes (II). The correct answer is (C). 7. A The question asks for the distance between X and Y. Use the vertical height to form two right triangles. Find the base of each triangle. The sum of the two bases will be the distance between X and Y. For reference, call the balloon point Z and the point on the ground directly below the balloon point W. Look at triangle XWZ. Angle X is 72.4°. WZ, which is opposite the angle X, is 54. The needed side is XW, which is adjacent to the angle X. Therefore, tan . Multiply both sides by a to get atan (72.4°) = 52. Divide both sides by tan (72.4°) to get . Now, do the same for triangle YWZ. Angle Y is 50.8°. WZ, which is opposite angle Y, is 54. The needed side is YW, which is adjacent to angle Y. Therefore, tan50.8° = 54. . Multiply both sides by a to get atan (50.8°) Divide both sides by tan (50.8°) to get . Add XW to WZ to get 17.13 + 44.04 = 61.17. The correct answer is (A). 8. B The question asks for the value of y2. Since , square both sides to get y2 = 342 − 302. Put the right side of the equation into a calculator to get 342 − 302 = 1,156 − 900 = 256. The correct answer is (B). 9. E Because there are variables in the choices, plug in. Pick coordinates for point A. Because the question involves distance, choose a point that can be used to make a Pythagorean triple. Let A be (x, y) = (3, 4). Point A’ is (3x, 3y) = (9, 12). The distance between A and the origin is c. Draw a segment vertically from A to the xaxis, forming a right triangle. The distance to the xaxis is 4, and the distance along the xaxis is 3. Therefore, this is a 3:4:5 right triangle, and c = 5. Do the same for A’. Draw a vertical line from (9, 12), forming a right triangle. The horizontal side has a length of 9, and the vertical side has a length of 12. Therefore, this is a 9:12:15 right triangle, and the distance from A’ to the origin is 15, which is the target. (Alternatively, use the Pythagorean Theorem to determine the hypotenuse of both triangles.) Plug c = 5 into each choice and eliminate any that aren’t 15. Choice (A) is Choice (B) is , so eliminate (A). , so eliminate (B). Choice (C) is 5, soeliminate (C). Choice (D) is 5 , so eliminate (D). Choice (E) is 3(5) = 15, so keep (E). The correct answer is (E). 10. D There are variables in the choices, so plug in. Let x = 3. Then, ≈ 15.389. This is the target. Go through the choices, one at a time and determine for which expression for q(x) does p(q(3)) ≈ 15.389. For (A), if q(3) = 32 − 2 = 7, then . Eliminate (A). For (B), if q(3) = 32 = 9, then Eliminate (B). For . (C), if q(3) = 3, then . Eliminate (C). For (D), if , then , so keep (D). For (E), if q(3) = then so eliminate (E). The correct answer is (D). 11. C The question asks for sin (90° − x). There are two possible approaches to this problem. One is to find the value of x by using the inverse cosine function. If cosx = 0.6, then take the inverse cosine of both sides to get x = cos−1(0.6) ≈ 53.13. Therefore, sin (90° − x) ≈ sin (90° − 53.13°) ≈ 0.6. Alternatively, use the identity cosx = sin (90° − x). Therefore, if cosx = 0.6, then sin (90° − x) = 0.6 Using either method, the correct answer is (C). 12. E In xyzcoordinates an equation with the graph x2 + y2 + z2 = r2 is a sphere with radius r and center at the origin. However, if this equation is not familiar, the question can still be answered using POE. Find points that satisfy this equation. Start with points (2, 0, 0), (0, 2, 0), and (0, 0, 2). Because there is more than one point, eliminate (A). These three points do not form a line, so eliminate (B). These points could make a circle, plane, or sphere, so Plug In more points. Try (−2, 0, 0), (0, −2, 0), and (0, 0, −2). These six points are not on the same plane, so eliminate (D). Since all points in any circle must be on the same plane, eliminate (C), as well. Only one choice remains. The correct answer is (E). 13. B The question asks for the xvalues at which g has vertical asymptotes. A function has a vertical asymptote for xvalues at which the denominator is 0 and the factor that makes the denominator equal to 0 cannot be canceled out with the numerator. Since the numerator of g cannot be factored, only worry about where the denominator is 0. Set x2 − 6x + 9 = 0. Factor the left side, finding two factors of 9 with a sum of −6. These are −3 and −3. Therefore, the factored form of the equation is (x − 3)(x − 3) = 0. Set both factors to 0 and solve. In both cases, the equation is x − 3 = 0, so add to both sides to get x = 3. The correct answer is (B). 14. C The question asks for the value of k, which is the constant term in the polynomial. The constant term represents the yintercept. According to the graph, the curve crosses the yaxis between 0 and 40. Only one choice is between these. The correct answer is (C). 15. C The question asks for cscx, which is equivalent to . Substitute the value of sinx given by the question to get csc . The correct answer is (C). 16. D The question asks for the average cost for each night, which is . There are variables in the choices, so plug in. Let n = 4. Since n represents the number of nights, let this be the denominator. Determine the total cost. Her stay at the hotel costs $80 per night for four nights for a total of 4 × $80 = $320. Furthermore, the threenight hotel stay for her friend costs 3 × $80 = $240. She must also pay airfare, which is $170. Therefore, the total cost is $320 + $240 + $170 = $730, and the average cost is . This is the target number. Plug n = 4 into each of the choices, and eliminate any that aren’t 182.5. Choice (A) is . Eliminate (A). Choice (B) is . Eliminate . Eliminate (B). (C). Choice Choice (C) is (D) is . Keep (D). Choice (E) is . Eliminate (E). The correct answer is (D). 17. A The set of points equidistant between two points is the perpendicular bisector of the segment whose endpoints are the two points. The segment with endpoints (0, 0) and (6, 0) lies on the line y = 0 and has midpoint (3, 0). Since a line in the form y = c, where c is a constant, is a horizontal line, the perpendicular line must be a perpendicular line the form x = k, where k is a constant. To be a bisector, the line must go through the midpoint, which is (3, 0), so the line is x = 3. Alternatively, sketch the two points and sketch each of the choices. Choices (B), (C), (D), and (E) all have points that are clearly closer to (0, 0) to (6, 0) and vice versa, so they can be eliminated. The correct answer is (A). 18. B A geometric series is one with nth term arn−1, where a represents the first term, and r represents the common ratio, i.e. the number by which each term must be multiplied to get the next term. If 0 < r < 1, then the sum of an infinite series can be found using the formula . The first term is , fill in this for a. To find the common ratio, set up the equation Multiply both sides by 9 to get . . Therefore, the sum is . Alternatively, find the sum of the known terms on a calculator This is close to . Using either method, the correct answer is (B). 19. D Simplify the inequality by combining like terms. Subtract a from both sides to get −b ≥ b. Add b to both sides to get 0 ≥ 2b. Divide both sides by 2 to get 0 ≥ b. This can be rewritten as b ≤ 0. The correct answer is (D). 20. E The question says must be true, so eliminate any choice that can be false. It is unknown whether g is increasing, decreasing, or neither, so attempt as many cases as is needed to eliminate four choices. Let g be an increasing function. Try g(x) = x. In this case, if g(m) >g(n), then m > n. Eliminate (C) and (D). Furthermore, it could be that m = 2 and n = 0. In this case, mn = 0, so eliminate (A) as well. Now try g(x) = −x. In this case, if g(m) > g(n), −m > −n. Divide both sides by −1 to get m <n. Eliminate (B). The correct answer is (E). 21. B The question involves percents, so plug in 100 as the total number of employees in the office. 60% of the employees commute over an hour each day on average. Since 60% of 100 is 60, exactly 60 employees commute over an hour each day on average. 25% of those 60 employees commute by train. Since 25% of 60 is , there are 15 employees who commute over an hour by train. The question asks for the probability that a randomly selected employee commutes over an hour on average by train. There are 15 employees who do this out of 100 total employees, so the probability is 22. E = 0.15. The correct answer is (B). The second smallest angle is the one opposite the second shortest side. The second shortest side is the 12. Mark the angle opposite the 12 as x°. The hypotenuse is 13. Therefore, sin . Since sinx = take sin−1 of both sides to get x ≈ 67.38. The question asks for the nearest degree, so the correct answer is (E). 23. D The question asks for the equation of the line perpendicular to y = 3x − 5. In the xyplane, perpendicular lines have slopes that are negative reciprocals. Find the slope of y = 3x − 5. For a line in the form y = mx + b, the slope is represented by m. Therefore, the slope of y = 3x − 5 must be 3. Find a choice that represents the equation of a line with a slope of − . Choice (E) is not the equation of a line, so eliminate (E). The remaining choices are in y = mx + b form, so select the choice with m = − . The correct answer is (D). 24. D The question asks for the range of g(x), which is a cosine function. The range of cos (x) is −1 ≤ cos (x) ≤ 1. Anything that’s inside the parentheses does not affect the range. The coefficient 3 only affects the period and not the range. Similarly, 7π only shifts the graph to the left and does not affect the range. Therefore, the range of cos (3x + 7π) is −1 ≤ cos (3x + 7π) ≤ 1. The coefficient 5 changes the amplitude, which is defined as half the distance from the maximum and minimum values of the cosine graph. Therefore, the range of 5cos (3x + 7π) is −5 ≤ 5cos (3x + 7π) ≤ 5. The constant −2 represents a downward shift, so subtract 2 from both the maximum and the minimum. Therefore, the range of g(x) is −7 ≤ g(x) ≤ 3. The correct answer is (D). 25. D The question asks for the list with the greatest standard deviation. The list with the greatest standard deviation is the list in which the numbers are farthest apart. The numbers in (D) have the greatest separation. The correct answer is (D). 26. E First, plug in the known values into the equation. The initial investment is $100, so plug in I = 100. The account needs to be worth $600, so plug in F = 600 to get 600 = 100e0.06y. Solve for y. Divide both sides by 100 to get 6 = e0.06y. Take the natural log of both sides to get ln (6) = 0.06y and 1.791759 = 0.06y. Divide both sides by 0.06 to get 29.86 = y, which is closest to 29.7. The correct answer is (E). 27. C The question asks for the location of θ on the xyplane. Use the unit circle. In the unit circle, cosθ is equivalent to the xcoordinate, and sinθ is equivalent to the y coordinate. If cosθ < 0, then x < 0. The xcoordinate is negative in quadrants II and III. Eliminate (A) and (D). The question also states that . If a fraction is positive, then the numerator and denominator must be of the same sign. Since cosθ is negative, so is sinθ. Thus, the ycoordinate must be negative. The x and ycoordinate are both negative in Quadrant III, so the correct answer is (C). 28. B The question asks for what point must be on the graph of g. The only point on the graph of g that is given is (4, 9). Therefore, g(4) = 9. Since g(−x) = −g(x), g(−4) = −g(4) = −9. Since g(−4) = −9, the point (−4, −9) must also be on the graph of g. The correct answer is (B). 29. A The question asks for what statement CANNOT be inferred from the given statement. Because the question says CANNOT, ignore the CANNOT and mark each choice as Y or N depending of whether it can be inferred. The given statement is a conditional statement in the form if p then q. In the original statement, p is a multiple of 10 and q is a multiple of 5. A conditional statement is logically equivalent to its contrapositive, which is in the form if not q, then not p. Therefore, the contrapositive of the given statement If a is a multiple of 10, then a is a multiple of 5 is If a is not a multiple of 5, then a is not a multiple of 10. Go through each choice one at a time. Choice (A) is in the form if q then p. A reversal of the order of the original condition statement cannot be assumed to be equivalent to the original. (For example, consider a = 15.) Mark this choice as N. Choice (B) is the contrapositive. The contrapositive is logically equivalent, so mark this choice with Y. Choice (C) rephrases an if p then q statements as an p implies q. This is logically equivalent, so mark this choice with Y. Choice (D) discusses a necessary condition. For a statement in the form if p then q, p is referred to as the sufficient condition and q is referred to as the necessary condition. Since (D) refers to q as the necessary condition, mark (D) with Y. Similarly, (E) refers to p as the sufficient condition, so mark (E) with Y. Four choices are marked as Y and one choice is marked with N, so select the choice marked with N. The correct answer is (A). 30. C The question asks for how many different arrangements of 8 flowers in a line. Draw 8 spaces in a line for the 8 flowers. Consider the number of possibilities for each space. For the first space, any of the 8 flowers can be chosen, so put an 8 in the first space. Once that flower is chosen, there are 7 possible flowers remaining for the second choice, so put a 7 in the second space. Similarly, there are 6 flowers remaining for the 3rd space, 5 for the 4th space, 4 for the 5th space, 3 for the 6th space, 2 for the 7th space, and 1 for the 8th space. Because the question asks for different orders, the order matters, so multiply the numbers in the spaces without doing any division to get . The correct answer is (C). 31. D The question asks for what value an expression approaches as x approaches 0. Plug in a value close to 0. Let x = 0.1. In a calculator, compute to get 2.098. Eliminate (A), (B), and (C) as they are not close to 2.098. To determine whether the answer could be (E), plug in a value for x that is slightly less than 0. Let x = −0.1. In a calculator, compute to get 1.898. This is still close to 2, so the function does approach 2 as x approaches 0. As an alternative, graph on a graphing calculator. The graph is a mostly smooth graph with a hole at (0, 2). Using either method, the correct answer is (D). 32. E The question asks for f(1). Replace x with 1 in f(x) to get f(1) = 7 − 5(1) = 2. Go through the choices and find the one equal to 2. In (A), f(−1) = 7 − 5(−1) = 12. Eliminate (A). In (B), f(0) = 7 − 5(0) = 7. Eliminate (B). In (C), . Eliminate (C). In (D), f(2) = 7 − 5(2) = 3. Eliminate (D). In (E), . Keep (E). The correct answer is (E). 33. B The question asks for the period of a tangent graph. The period of a tangent graph in the form y = Atan (Bx + C), the period is equal to the expression period is . In this equation B = 2π, so the . Alternatively, graph the equation and find the distance between zeroes. This distance is equal to the period. The correct answer is (B). 34. D The question asks for how long it would take Katherine to get from Point X to Point Y along Maple Street and Elm Street. Because this is a rate question, use the formula d = rt, with d representing distance, r representing rate, and t representing time. According to the question, the rate is 15 kilometers per hour, so substitute r = 15 to get d = 15t. To get the time, find the distance. According to the question, the straightline distance from Point X to Point Y is 40 kilometers. However, the question also says that she bikes along Maple Street and Elm Street, so don’t use the straightline distance for d. Instead use the sum of the distances from Point X to the intersection and from the intersection to Point Y for d. The distance from the Point X to the intersection is 10. To find the distance from intersection to Point Y, use the Pythagorean Theorem: a2 + b2 = c2. Draw the straightline distance from Point X to Point Y and label it 40. Since 40 is the hypotenuse, plug in a = 10 and c = 40 to get 102 + b2 = 402. Simplify to get 100 +b2 = 1600. Subtract 100 from both sides to get b2 = 1500. Take the square root of both sides to get b ≈ 38.7298. To get the total distance, add this to the distance from Point X to the intersection to get d = 10 + 38.7298 = 48.7298. Plug this into the equation to get 48.7298 = 15t. Divide both sides by 15 to get t ≈ 3.2487. This is the time in hours. Because it is between 3 and 4 hours, eliminate (A), (B), and (C). The remaining two choices only differ by the remainder in minutes. To get the remainder in minutes, set up the proportion . Cross multiply to get x ≈ 15 minutes. Therefore, the total time is 3 hours and 15 minutes. The correct answer is (D). 35. E The question asks for what could equal g(x). The choices are in factored form. In the factored form of a polynomial, each factor is in the form (x −r) where r is one of the roots, so find the roots. The roots of an equation are the values of x at which the value of the function is 0. According to the table g(x) = 0, when x = −2, 1, and 2. Therefore, the factors (x + 2), (x − 1), and (x − 2) must be included in the equation of the function. Eliminate (A) and (C). Plug the remaining points into the answer choices. Begin with f(0) = 2. In (B), f(0) = (0 − 2)(0 − 1)(0 + 2)(0 + 3) = 12, so eliminate (B). In (D), f(0) = (0 − 3)(0 − 2)(0 − 1)(0 + 2) = −12, so eliminate (D). In (E), f(0) = (0 − 2)(0 − 1) , so keep (E). The correct answer is (E). 36. E The question asks for what could NOT be a factor of m, an integer whose only prime factors are 2, 3, 5, and 13. A factor of m must, by definition, divide m evenly with no remainder. If a number is to divide m evenly, every factor of the number must also be a factor of m. Therefore, if a number has any prime factor other than 2, 3, 5, or 13, the number cannot be a factor of m. Find the prime factors of each choice. Because the question says NOT, ignore the NOT. Instead mark a choice with Y if it could be a factor and N if it could not be. The prime factors of 6 are 2 and 3. Since 2 and 3 are prime factors of m, this could be a factor, so mark (A) with Y. The prime factors of 9 are 3 and 3. Since 3 is a prime factor of m, this could be a factor, so mark (B) with Y. The prime factors of 12 are 2, 2, and 3. Since 2 and 3 are prime factors of m, this could be a factor, so mark (C) with Y. The prime factors of 26 are 2 and 13. Since 2 and 13 are prime factors of m, this could be a factor, so mark (D) with Y. The prime factors of 35 are 5 and 7. Since 7 is not a prime factor of m, this cannot be a factor, so mark (E) with N. Four choices are marked with Y and one choice is marked with N, so select the choice marked with N. The correct answer is (E). 37. A The question asks for the value of x. There are values in the choices, so PITA. Start with (C). If x = 0.328, then cosx ≈ 0.947 and 4sinx ≈ 1.289. These are not equal, so eliminate (C). Determine whether the answer must be greater or less than 0.328. Use the unit circle. As the angle goes from 0 to on the unit circle, then xcoordinate decreases and ycoordinate increases. The xcoordinate corresponds with cosine, and the y coordinate corresponds with sine. Therefore, as the angle increases, cosine decreases and sine increase. Since cos (0.328) < 4sin (0.328), the angle must decrease, so that the cosine increases and the sine decreases. Eliminate (D) and (E). Try (B). If x = 0.250, then cosx ≈ 0.969 and 4sinx ≈ 0.990. These are not equal, so eliminate (B). Only one choice remains but try (A) to be sure. If x = 0.245, the cosx ≈ 0.970 and 4sinx ≈ 0.970. Although a calculator indicates that the values are not exactly equal, the answer choices on the test are rounded, so if they are extremely close, they can be taken to be equal. Therefore, the correct answer is (A). 38. D The question asks for the value of g−1(15). The function g−1 is defined as the inverse of g. For inverse functions, if g(x) = y, then g−1(y) = x. Therefore, y = 15 and the question asks for the value of x. Since the question asks for the value of x and there are numbers in the choices, PITA. Plug the choices in for x and eliminate any for which f(x) is not 15. Start with (C). If x = 3.17, then g(x) = g(3.17) ≈ 11.944. This is not 15, so eliminate (C). The answer must be greater, so eliminate (A) and (B), as well. Try (D). If x = 5, then g(x) = g(5) = 15. This matches the target, so the correct answer is (D). 39. C The question asks for the 11th term of the Triangular Number Sequence. To find a term in a recursive sequence, start at the beginning. The definition states that T1 = 1 and that Tn = Tn− 1 + n for n > 1. Therefore, T2 = T1 + 2 = 1 + 2 = 3. Continue to T11. T3 = T2 + 3 = 3 + 3 = 6. T4 = T3 + 4 = 6 + 4 = 10. T5 = T4 + 5 = 10 + 5 = 15. T6 = T5 + 6 = 15 + 6 = 21. T7 = T6 + 7 = 21 + 7 = 28. T8 = T7 + 8 = 28 + 8 = 36. T9 = T7 + 9 = 36 + 9 = 45. T10 = T9 + 10 = 45 + 10 = 55. T11 = T10 + 11 = 55 + 11 = 66. The correct answer is (C). 40. E The question asks which statements are true about the function f(x) = x3 + x2 − 16x + 12 are true. If a graphing calculator is available, graph the function. The graph crosses the xaxis three times, so statement I is true. Also, the graph has a relative minimum at (2, −8) and has no points for which y < −8 on the positive side of the xaxis. Therefore, II and III are also true. If no graphing calculator is available, then factor to determine the number of solutions. One method is to test factors of 12 for a solution. The factors of 12 are 1, 2, 3, 4, 6, and 12. Try x = 1. Since f(1) = 13 + 12 − 16(1) + 12 = −2, x = 1 is not a solution. Try x = 2. Since f(2) = 23 + 22 − 16(2) + 12 = −8, x = 2 is not a solution. Since f(3) = 33 + 32 − 16(3) + 12 = 0, x = 3 is a solution and (x − 3) is one factor. Rewrite x3 +x2 − 16x + 12 in a way that makes it easy to factor (x − 3). First, rewrite it as x3 − 3x2 + 3x2 + x2 − 16x + 12, and factor the first two terms to get x2(x − 3) + 4x2 − 16x + 12. Now, rewrite it as x2(x − 3) + 4x2 − 12x + 12x − 16x + 12, and factor 4x2 − 12x to get x2(x − 3) + 4x(x − 3) − 4x + 12. Now factor −4x + 12 to get x2(x − 3) + 4x(x − 3) − 4(x − 3). Factor (x − 3) to get (x − 3)(x2 + 4x − 4). Now determine the number of factors of x2 − 4x + 4. To determine the number of factors of a quadratic in the form ax2 + bx + c, use the discriminant: b2 − 4ac. If the discriminant is positive, there are two real solutions. If the discriminant is 0, there is one real solution. If the discriminant is negative, there are no real solutions. In the quadratic x2 + 4x − 4, a = 1, b = 4, and c = −4, so b2 − 4ac = 42 − 4(1)(−4) = 32 > 0. Since the discriminant is positive, x2 + 4x − 4 has two solutions and (x − 3)(x2 + 4x − 4) has three solutions. Thus, (I) is true. Eliminate (B) and (D). Test (II), which says that f(x) ≥ −8, for all x ≥ 0. Set up x3 + x2 − 16x + 12 ≥ −8. Get one side equal to 0. Add 8 to both sides to get x3 + x2 − 16x + 20 ≥ 0. Similarly, factor the polynomial on the right by testing the factors of 20: 1, 2, 4, 5, 10 and 20. Since 13 + 12 − 16(1) + 20 = 6, x = 1 is not a solution. Since 23 + 22 − 16(2) + 20 = 0, x = 2 is a solution, so (x − 2) is a factor. Rewrite x3 + x2 − 16x + 20 as x3 − 2x2 + 2x2 + x2 − 16x + 20 = x2(x − 2) + 3x2 − 16x + 20 = x2(x − 2) + 3x2 − 6x + 6x − 16x + 20 = x2(x − 2) + 3x(x − 2) − 10x + 20 = x2(x − 2) + 3x(x − 2) − 10(x − 2) = (x − 2) (x2 + 3x − 10). Factor to get (x − 2)(x − 2)(x + 5) = (x − 2)2(x + 5). Therefore, the expression x3 + x2 − 16x + 20 = 0 when x = 2 and x = −5. The statement only refers to what happens when x ≥ 0, so ignore x = −5. Since x3 + x2− 16x + 20 = 0, when x = 2, x3 + x2 − 16x + 12 = f(x) = −8, when x = 2. Determine what happens to the left and right of x = 2. If x = 1, then f(1) = 13 + 12 − 16(1) + 12 = −2 ≥ −8. If x = 3, then f(3) = 33 + 32 − 16(3) + 12 = 0 ≥ −8. Therefore f(x) ≥ −8, whenever x ≥ 0, so (II) is true. Eliminate (A) and (C). Only one choice remains. The correct answer is (E). 41. D The question asks which of the following could be a portion of the graph of gh, the product of the graphs of g and h. Look at the graphs in pieces. When x is negative, the graphs of g and h are both negative. Therefore, the product must be positive. Eliminate (A) and (C), which are positive when x is negative. If x is positive, g is positive but h is negative, so the product must be negative. Eliminate (B), which is positive when x is positive. If x = 0, the f and g are both 0, so the product is 0, and the graph of the product must go through the origin. Eliminate (E), which does not go through the origin. The correct answer is (D). 42. D The question asks what set of real numbers makes the equation true. Plug In values of y. Try y = 0. If y = 0, then the equation is true. Eliminate (B) and (C), which indicate that 0 should not be included. Plug in a positive number. Try y = 2. If y = 2, then the equation is true. Eliminate (E), which says that 2 should not be included. Now, try a negative. Try y = −2 If y = −2, then the equation is false. Eliminate (A), which says that −2 should be included. Only one choice remains. The correct answer is (D). 43. D The question asks for sinx. The triangle is not necessarily a right triangle. In nonright triangles, use the law of sines: . The side opposite the 60° angle is 12 and the side opposite the x° angle is 5. Plug these into the formula to get . Cross multiply to get 5sin (60) = 12sin (x). Use the 306090 right triangle to get sin (60) . Divide both sides by 12 to get . Therefor, . The correct answer is (D). 44. C The question asks for the longest segment that can be drawn between two vertices of the solid. The longest segments are always the diagonals that go through the center of the solid. To find the length of one of them, use the threedimensional version of the Pythagorean Theorem: a2 + b2 + c2 =d2, where a, b, and c represent the dimensions of the solid and d represents the diagonal. Plug in the dimensions to get 62 + 32 + 22 = d2. Simplify the left side to get 49 = d2. Take the square root of both sides to get 7 =d. The correct answer is (C). 45. D The question asks for logn50. The question also uses variables to represent logn2 and logn5, so put logn50 into those terms. Notice that 2 and 5 are the prime factors of 50. Since 50 = 2 × 52, logn50 = logn(2 × 52). A product within a log is equal to the sum of the logs. Therefore, logn(2 × 52) = logn(2) + logn(52). An exponent within a log can be moved in front of the log as a coefficient. Therefore, logn(2) + logn(52) = logn(2) + 2logn(5). Since logn2 = a, and logn5 =b, logn(2) + 2logn(5) = a + 2b. Alternatively, plug in n = 10 and use a calculator to get log 2 ≈ 0.301, log 5 ≈ 0.699, and log 50 = 1.699. Therefore, a = 0.301, b = 0.699, and the target is 1.699. Go through the choices and eliminate any that aren’t 1.699. Choice (A) is 0.301 + 0.699 = 1, so eliminate (A). Choice (B) is 0.301 + 0.6992 = 0.790, so eliminate (B). Choice (C) is (0.301)(0.699)2 = 0.147, so eliminate (C). Choice (D) is 0.301 + 2(0.699) = 1.699, so keep (D). Choice (E) is 0.301 + 5(0.699) = 3.796, so eliminate (E). The correct answer is (D). 46. E The question asks for tanx. There are variables in the choices, so plug in. Since cosx = a, plug in for x and find a. To make sure the numbers are easy, plug in x = to 60°. If x = , then tanx. Since x = radians, which is equivalent . The question asks for , tanx = tan = tan60° = ≈ 1.73. This is the target. Go through the choices and eliminate any that aren’t equal to so . Choice (A) is eliminate . This is not (A). Choice (B) . This is not eliminate (B). Choice (C) is is not , so eliminate This is , so keep (E). is , so . This (C).Choice (D) . This is not eliminate (D). Choice (E) is , is , so . 47. A The question asks for the shift in the graph. There are variables in the choices, so plug in. Since c is a constant greater than 5, plug in c = 6. Then, the equation is y = x2 + 4x + 6. To determine the shift, find the vertex by getting the equation in vertex form. Complete the square. Set y = 0 to get 0 = x2 + 4x + 6. Subtract 6 to move the constant to the other side to get −6 = x2 + 4x. Add the square of half the coefficient on x to both sides. The coefficient is 4, half of 4 is 2, and the square of 2 is 4. Add 4 to both sides to get −2 = x2 + 4x + 4. Factor the right side to get −2 = (x + 2)2. Add 2 to both sides to get 0 = (x + 2)2 + 2. Replace 0 with y to get y = (x + 2)2 + 2. Now, the equation is in the form y = (x − h)2 + k, where (h, k) represents the vertex of the parabola. Therefore, the vertex is (−2, 2). Since the vertex of y = x2 is (0, 0), the graph shifts to the left 2. Eliminate (B), (C), and (E). The graph also shifts up 2 units. Since c = 6, (A) says that the graph shifts up 6 − 4 = 2 units, while (D) says that the graph shifts 6 + 4 = 10. Eliminate (D). The correct answer is (A). 48. B The question is asking for the percent that the side of the base must be increased. Since the question is asking for a percent of some unknown value, the hidden plug in approach is the best to use here. The volume formula for a rectangular pyramid is V = lwh, where l and w represent the length and width, respectively, of the base, and h represents the height of the pyramid. First, plug in numbers just to calculate a starting volume. Since the base of the pyramid is a square, the length and width will be the same. Let l = 2, w = 2, and h = 3. Plugging into the formula yields: V = (2)(2)(3) = 4. Next, calculate the increased volume of the pyramid: 28% of 4 = 0.28 × 4 = 1.12. The volume that is 28% greater is 4 + 1.12 = 5.12. The question states that the height is increased by 12%, so calculate the new height: 12% of 3 = 0.12 × 3 = 0.36. The height that is 12% greater is 3 + 0.36 = 3.36. Finally, plug the new values calculated for volume and height into the formula to figure out the new values for length and width, which will both just equal x since the values are equal: (5.12) = (x)(x)(3.36). Simplifying results in 5.12 = 1.12x2. Divide both sides by 1.12 to get 4.571428571 = x2, and x = 2.138089935. The original side length was 2, and the question is asking for the percent the side length needs to be increased. Use the percent change formula of to find the percent: , which is closest to 7%. The correct answer is (B). 49. D The question asks which statements must be true, and the statements involve the dimensions and existence of matrix products. In order for a matrix product to exist, the number of columns of the first matrix must equal the number of rows of the second matrix. For the product XY, Matrix X has b columns and Matrix Y has b rows. Therefore, the product does exist. The number of dimensions matches the rows of the first matrix and the number of columns of the second. Since Matrix X has a rows and Matrix Y has c columns, the product XY has dimensions a × c. Therefore, (I) is true and (II) is false. Eliminate (B), (C), and (E). For the product YX, Matrix Y has c columns and Matrix X has a rows. Because these are not equal, the product YX does not exist. Therefore, (III) is true. Eliminate (A). The correct answer is (D). 50. E The question asks what could be iz. Plug in for z, which is a complex number in Quadrant IV. A complex number in Quadrant IV is in the form a +bi, where a > 0 and b < 0. Let z = 2 − 3i. Multiply both sides by i to get iz = 2i − 3i2. Since i2 = −1, iz = 2i − 3(−1) = 3 + 2i. Since both a and b are positive, iz must be in Quadrant I, and therefore at point E. The correct answer is (E). HOW TO SCORE PRACTICE TEST 1 When you take the real exam, the proctors will collect your test booklet and bubble sheet and send your bubble sheet to a processing center where a computer looks at the pattern of filledin ovals on your bubble sheet and gives you a score. We couldn’t include even a small computer with this book, so we are providing this more primitive way of scoring your exam. Determining Your Score STEP 1 Using the answer key, determine how many questions you got right and how many you got wrong on the test. Remember: Questions that you do not answer don’t count as either right answers or wrong answers. STEP 2 List the number of right answers here. (A) ______ STEP 3 List the number of wrong answers here. Now divide that number by 4. (Use a calculator if you’re feeling particularly lazy.) (B) ______ ÷ 4 = (C) ______ STEP 4 Subtract the number of wrong answers divided by 4 from the number of correct answers. Round this score to the nearest whole number. This is your raw score. (A) ______− (C) ______= ______ STEP 5 To determine your real score, take the number from Step 4 and look it up in the left column of the Score Conversion Table on the next page; the corresponding score on the right is your score on the exam. PRACTICE TEST 1 SCORE CONVERSION TABLE Click here to download and print the PDF version of this table. Part III Content Review 5 6 7 8 9 10 11 12 Algebra Fundamentals Plane and Solid Geometry Coordinate Geometry Trigonometry Functions Statistics and Sets Miscellaneous Chapter 5 Algebra On the SAT Subject Test in Math 2 you will often be asked to solve for a variable. While you likely think of yourself as an algebra whiz, ETS has many tricks up its sleeve to trip you up. Fear not! We’ll show you some tricks and techniques to avoid falling for ETS’s traps. In addition, we’ll cover solving for x, inequalities, factoring, simultaneous equations, quadratics, and more! (You may be wondering why algebra comes before “fundamentals.” As you will see in the next chapter, many of our techniques which simplify algebra problems will also simplify problems about fundamental math concepts. Trust us!) ALGEBRA ON THE SAT SUBJECT TEST IN MATH 2 Algebra questions will make up approximately 20 percent of the SAT Subject Test in Math 2. Many of these questions can be best answered using the algebra rules reviewed in this chapter. Others are best approached using some of the testtaking techniques discussed in Chapter 2. Definitions Here are some algebraic terms that will appear on the SAT Subject Test in Math 2. Make sure you’re familiar with them. If the meaning of any of these vocabulary words keeps slipping your mind, add those words to your flash cards. Variable An unknown quantity in an equation represented by a letter (usually from the end of the alphabet), for example, x, y, or z. Constant An unchanging numerical quantity—either a number or a letter that represents a number (usually from the beginning of the alphabet), for example, 5, 7.31, a, b, or k. Term An algebraic unit consisting of constants and variables including any operation other than addition and subtraction such as 5x or 9x2. Coefficient In a term, the constant before the variable. In ax2, a is the coefficient. In 7x, 7 is the coefficient. Polynomial An algebraic expression consisting of more than one term joined by addition or subtraction. For example, x2 − 3x2 + 4x − 5 is a polynomial with four terms. Binomial A polynomial with exactly two terms, such as (x − 5). Quadratic A quadratic expression is a polynomial with one variable whose largest exponent is a 2, for example, x2 − 5x + 6 or x2+ 4. Root A root of a polynomial is a value of the variable that makes the polynomial equal to zero. More generally, the roots of an equation are the values that make the equation true. Roots are also known as zeros, solutions, and xintercepts. Degree The greatest exponent on a variable in a polynomial or function. For example, f(x) = 3x4 + 3x − 2 is a function of the fourth degree. SOLVING EQUATIONS Some questions on the SAT Subject Test in Math 2 will require you to solve a simple algebraic equation. These questions often present you with an algebraic equation hidden in a word problem. Setting up an equation from the information in the problem is the first step to finding a solution and is the step where careless mistakes are often made. The translation chart on this page is very useful for setting up equations from information given in English. An algebraic equation is an equation that contains at least one unknown —a variable. “Solving” for an unknown means figuring out its value. Generally, the way to solve for an unknown is to isolate the variable—that is, manipulate the equation until the unknown is alone on one side of the equal sign. Whatever’s on the other side of the equal sign is the value of the unknown. Take a look at this example. 5(3x3 − 16) − 22 = 18 In this equation, x is the unknown. To solve for x, you need to get x alone. You isolate x by undoing everything that’s being done to x in the equation. If x is being squared, you need to take a square root; if x is being multiplied by 3, you need to divide by 3; if x is being decreased by 4, you need to add 4, and so on. The trick is to do these things in the right order. Basically, you should follow PEMDAS in reverse. Start by undoing addition and subtraction, then multiplication and division, then exponents and roots, and, last, what’s in parentheses. The other thing to remember is that any time you do something to one side of an equation, you’ve got to do it to the other side also. Otherwise you’d be changing the equation, and you’re trying to rearrange it, not change it. In this example, you’d start by undoing the subtraction. Then undo the multiplication by 5, saving what’s in the parentheses for last. Once you’ve gotten down to what’s in the parentheses, follow PEMDAS in reverse again—first the subtraction, then the multiplication, and the exponent last. At this point, you’ve solved the equation. You have found that the value of x must be 2. Another way of saying this is that 2 is the root of the equation 5(3x3 − 16) − 22 = 18. Equations containing exponents may have more than one root. Equations which have more than one root are common when you have an equation with an even degree. For example: 3x4 = 48 First, divide both sides by 3: x4 = 16 Then take the fourth root of each side. But, because you’re taking an even root of the equation, you may have a positive OR a negative value for x: Solving Equations with Radicals To solve equations with radicals, you work the equation the same way you work other equations. In order to eliminate the radical, you take both sides to the power of that radical. For example: Start by subtracting 2 from both sides, then divide both sides by 3: Next, square both sides, then divide by 4: 4x = 4; x = 1 Note that you do not get both a positive and negative root when you are working with radicals. Radical equations will have one solution. Solving Equations with Absolute Value The rules for solving equations with absolute value are the same. The only difference is that, because what’s inside the absolute value signs can be positive or negative, you’re solving for two different results. Let’s look at an example: 20. x −2 = 17 Now we know that either (x − 2) is a negative number or a nonnegative number. When a number is negative, the absolute value makes it the inverse, or multiplies it by −1 to yield a positive result. If the number is positive, it remains the same after being sent through the absolute value machine. So when we remove the absolute value bars, we’re left with two different equations: x − 2 = 17 or x − 2 = −17 Now simply solve both equations: And that’s all there is to it! DRILL 1: SOLVING EQUATIONS Practice solving equations in the following examples. Remember that some equations may have more than one root. The answers can be found in Part IV. 1. If , then x = 2. If n2 = 5n, then n = 3. If 4. If , then a = , then s = 5. If , then x = 6. If 2m + 5=23, then m= 7. If , then r = 8. If , then x = 9. If and y ≠ 0, then y= FACTORING AND DISTRIBUTING When manipulating algebraic equations, you’ll need to use the tools of factoring and distributing. These are simply ways of rearranging equations to make them easier to work with. Factoring Factoring simply means finding some factor that is in every term of an expression and “pulling it out.” By “pulling it out,” we mean dividing each individual term by that factor, and then placing the whole expression in parentheses with that factor on the outside. Here’s an example: x3 − 5x2 + 6x = 0 On the left side of this equation, every term contains at least one x—that is, x is a factor of every term in the expression. That means you can factor out an x: x3 − 5x2 + 6x = 0 x(x2 − 5x + 6) = 0 The new expression has exactly the same value as the old one; it’s just written differently, in a way that might make your calculations easier. Numbers as well as variables can be factored out, as seen in the example below. 11x2 + 88x + 176 = 0 This equation is, at first glance, a bit of a headache. It’d be nice to get rid of that coefficient in front of the x2 term. In a case like this, check the other terms and see if they share a factor. In fact, in this equation, every term on the left side is a multiple of 11. Because 11 is a factor of each term, you can pull it out: 11x2 + 88x +176 = 0 11(x2 + 8x +16) = 0 x2 + 8x +16 = 0 (x + 4)2 = 0 x = −4 As you can see, factoring can make an equation easier to solve. Distributing Distributing is factoring in reverse. When an entire expression in parentheses is being multiplied by some factor, you can “distribute” the factor into each term, and get rid of the parentheses. For example: 3x(4 + 2x) = 6x2 + 36 On the left side of this equation the parentheses make it difficult to combine terms and simplify the equation. You can get rid of the parentheses by distributing. And suddenly, the equation is much easier to solve. DRILL 2: FACTORING AND DISTRIBUTING Practice a little factoring and distributing in the following examples, and keep an eye out for equations that could be simplified by this kind of rearrangement. The answers can be found in Part IV. 3. If (11x)(50) + (50x)(29) = 4,000, then x = (A) 2,000 (B) 200 (C) 20 (D) 2 (E) 0.2 5. If ab ≠ 0; (A) −3 (B) −2 (C) 0 (D) 1 (E) 3 22. If x ≠ −1, (A) 4x2 (B) x2 (C) 4x (D) x (E) 4 30. If x ≠ 0, −1, then (x5 + 2x4 +x3)−1 = (A) (B) −x3(x+1)2 (C) (D) (E) PLUGGING IN Plugging In is a technique for shortcutting algebra questions. It works on a certain class of algebra questions in which relationships are defined, but no real numbers are introduced. For example: 1. The use of a neighborhood car wash costs n dollars for a membership and p cents for each wash. If a membership includes a bonus of 4 free washes, which of the following reflects the cost, in dollars, of getting a membership at the car wash and washing a car q times, if q is greater than 4 ? (A) 100n + pq − 4p (B) n + 100pq − 25p (C) n + pq − (D) n + + (E) n + − Here’s How to Crack It In this problem you see a lot of variables in the question and in the answer choices. That’s a big clue! When you see variables in the answer choices, PLUG IN! Let’s try Plugging In with this problem. We’ll start with n, the membership fee. Plug In an easy number like 3, so that a membership costs $3.00. Then, Plug In a number for p, the charge per wash. Since this number is in cents, and we’ll need to convert it to dollars in the answers, choose a number that can be converted easily to dollars, like 200. Let’s make p = 200, so a wash costs $2.00. To Number or Not to Number? Let’s say you walk into a candy store. The store is selling certain pieces of candy for 5 cents and 10 cents each. You want to get 3 pieces of the 5 cent candy and 6 pieces of the 10 cent candy. You give the cashier a $5 bill. What’s your change? Ok, now let’s say you walk into a candy store run by ETS. This store is selling certain pieces of candy for x cents and y cents each. You want to get m pieces of the x cent candy and n pieces of the y cent candy. You give the cashier a $z bill. What’s your change? Which problem would be easier to solve? The one with the numbers! Numbers make everything easier. So why bother with variables when you don’t have to? Last, let’s say that q, the number of washes, is 5. That’s as easy as it gets. With 4 free washes, you’re paying for only 1. Then, just work out the answer to the question using your numbers. How much does it cost for a membership and 5 washes? Well, that’s $3.00 for a membership, 4 washes free, and 1 wash for $2.00. The total is $5.00. That means that if you plug your numbers into the answer choices, the right answer should give you 5. We call that your target number—the number you are looking for in the answer choices. Put a double circle around your target number, so that it stands out from all the other numbers you’ve written down. It looks like a bull’seye that you’re trying to hit: When you plug n = 3, p = 200, and q = 5 into the answer choices, the only answer choice that gives you 5 is (D). That means you’ve hit your target number, and you’re done. A Big Clue There will be times when ETS will give you questions that include variables and the phrase “in terms of” (for example, “in terms of x”). This is a big clue that you can Plug In. Cross off the phrase “in terms of x,” because you don’t need it to solve the problem. Let’s look at a problem without variables: 3. The size of an art collection is tripled, and then 70 percent of the collection is sold. Acquisitions then increase the size of the collection by 10 percent. The size of the art collection is then what percent of its size before these three changes? (A) 240% (B) 210% (C) 111% (D) 99% (E) 21% Here’s How to Crack It Here’s another question in which you aren’t given numbers. In this case, you don’t know the original size of the art collection. Instead of variables, though, the question and answers contain percents. This is another sign that you can Plug In whatever numbers you like. Because you’re working with percentages, 100 is a good number to Plug In—it’ll make your math easier. You start with a collection of 100 items. It’s tripled, meaning it increases to 300. Then it’s decreased by 70%. That’s a decrease of 210, so the collection’s size decreases to 90. Then, finally, it increases by 10%. That’s an increase of 9, for a final collection size of 99. Since the collection began at 100, it’s now at 99% of its original size. The answer is (D). It doesn’t matter what number you choose for the original size of the collection— you’ll always get the right answer. The trick to choosing numbers is picking ones that make your math easier. Take a look at one last (challenging) problem: 50. If x dollars are invested at y percent annual interest, and 0 < x < 10,000 and y > 0, then in how many years will the value of the investment equal $10,000? (A) logx + log (1+0.01y) (B) (C) (D) (E) Here’s How to Crack It The answer choices with logarithms make this problem look extra intimidating. This problem deals with repeated percent change, but the variables in the question mean you can Plug In numbers to make life easier. If you make y = 100 then each year the value of the investment doubles. If you make x = 2,500, then after one year the investment is worth 5,000 and after two years the investment is worth 10,000. 2 is your target; circle it. Now you can plug in 2,500 for x and 100 for y in each answer, eliminating answers that do not equal 2. The only answer that equals 2 is (E). Remember to Plug In! 1. See variables in the question and answers? You can Plug In! 2. Assign a numerical value to each variable. Pick numbers that make the math easy. 3. Work the problem until you get a numerical answer. Circle it twice; that’s the target. 4. Plug In your numbers to each answer choice, eliminating every answer which doesn’t match your target. Always check every answer choice with variables! The idea behind Plugging In is that if these relationships are true, then it doesn’t matter what numbers you put into the question; you’ll always arrive at the same answer choice. So the easiest way to get through the question is to Plug In easy numbers, follow them through the question, and see which answer choice they lead you to. Occasionally, more than one answer choice will produce the correct answer. This often occurs when the question is looking for an exception, or something that “must be true.” When that happens, eliminate the answer choices that didn’t work out, and Plug In some different kinds of numbers. Some numbers you might try are odd and even integers, positive and negative numbers, fractions, 0, 1 or −1, and really big or really small numbers, like 1,000 or −1,000. The new numbers will produce a new target number. Use this new target number to eliminate the remaining incorrect answer choices. You will rarely have to Plug In more than two sets of numbers. When using Plugging In, keep a few simple rules in mind: • Avoid Plugging In 1 or 0, which often makes more than one answer choice produce the same number. For the same reason, avoid Plugging In numbers that appear in the answer choices— they’re more likely to cause several answer choices to produce your target number. • Plug In numbers that make your math easy—2, 3, and 5 are good choices in ordinary algebra. Multiples of 100 are good in percentage questions, and multiples of 60 are good in questions dealing with seconds, minutes, and hours. Plugging In can be an incredibly useful technique. By Plugging In numbers, you’re checking your math as you do the problem. When you use algebra, it takes an extra step to check your work with numbers. Also, there are fewer chances to mess up when you Plug In. And you can Plug In even when you don’t know how to set up an algebraic equation. Plugging In is often safer because ETS designs the answer choices so that, if you mess up the algebra, your result will be one of the wrong answers. When your answer matches one of the choices, you think it must be right. Very tempting. Furthermore, all of the answer choices look very similar, algebraically. This is how ETS camouflages correct answers. But when you Plug In, the answers often look very different. Often you’ll be able to approximate to eliminate numbers that are obviously too big or too small, without doing a lot of calculation, and that will save you lots of time! It is strongly recommended that you become comfortable with Plugging In. As you will see throughout this book, Plugging In often turns a challenging problem into a piece of cake. Even